Download Chapter 5 Thermochemistry Notes File

Chapter 5, Brown, LeMay, Bursten
Thermochemistry
Thermochemistry—(heat power)--the study of energy and its transformations
--includes concepts of energy, work, and heat.
I. Energy
Kinetic energy – energy of motion
-- KE = ½ mv2
-- atoms and molecules have mass and are in motion
Potential energy – stored energy that results from the attractions and repulsions an
object experiences in relation to other object
-- in chemistry, these attractions and repulsions are the
electrostatic forces between charged particles. (electron near a
proton has PE due to their attraction)
Chemical energy – due to the potential energy stored in the arrangements of the
atoms of the substance
Thermal energy – energy possessed by a substance because of its temperature
associated with the kinetic energy of the molecules
Joule – SI unit for energy = 1 kg-m2/s2
Calorie = the amount of energy required to raise the temperature of 1 gram of
water 1°C
1 calorie = 4.184 Joule
A closed system can exchange energy but not matter with its surroundings.
In lab, the chemicals are the system, the container and everything beyond it are the
surroundings. No exchange of matter with its surroundings, but an exchange of energy with
the surroundings in the form of work and heat.
Energy can be transferred (1) to cause the motion of an object against a force or
(2) to cause a temperature change.
Force = any kind of push or pull exerted on an object. (Newtons—kg-m/s2)
Work = energy used to cause an object to move against a force (Joules)
Work = F x d
(Work = m x g x d)
Heat = energy transferred from a hotter object to a colder one.
Energy = the capacity to do work or to transfer heat.
II. The First Law of Thermodynamics—energy can be neither created or destroyed:
ENERGY IS CONSERVED.
Internal energy – the sum of all kinetic and potential energy of all the components
of the system
= symbol E
-- the change in internal energy is
∆E = Efinal – Eintitial
-- can find ∆E even when Efinal and Einitial are not known.
-- ∆E has three parts, the number, the unit for magnitude, the sign
for direction
-- +∆E then Efinal > Einitial , so system has gained energy from surroundings
-- –∆E then Efinal < Einitial, so system has lost energy to the surroundings
-- Einitial refers to the reactants
-- Efinal refers to the products.
-- Energy is exchanged with surroundings as either heat or work.
So ∆E = q + w where q = heat added or liberated to system AND
w = work done on or by system
-- q is + if heat is transferred to the system from the surroundings
-- w is + if work is done to system by surroundings
THINK ABOUT ALL THE POSSIBLE COMBINATIONS FOR q AND w
AND THE RESULTING ∆E VALUE.
The hydrogen and oxygen gases in a cylinder with a piston are ignited. As the reaction occurs, the
system loses 1150 J of heat to the surroundings. The reaction also causes the piston to rise as the hot
gases expand. The expanding gas does 480 J of work on the surrounding as it pushes against the
atmosphere. What is the change in the internal energy of the system?
q = ?, h = ? ∆ = ?
Endothermic – a process in which the system absorbs heat from the surroundings (feels
cold)
Exothermic – a process in which the system releases heat to the surroundings (feels hot)
Energy is an EXTENSIVE property since the total internal energy is proportional to the
total quantity of matter in the system.
Energy is affected by temperature and pressure.
State function—a property of a system that is determined by specifying its condition or
its state (in terms of temperature, pressure, location, etc.)
-- value only depends on the present condition (not its history or how it got
there)
-- so internal energy ∆E is a state function, but q and h are NOT state
functions (q and h depend on the way in which the change is made)
III. Enthalpy
Most physical and chemical changes are done at constant pressure (earth’s atmosphere), so
most of the energy gained or lost by a system is in the form of heat.
Enthalpy = the heat absorbed or lost under constant pressure.
-- symbol = H
-- a state function (depends only on the initial and final states of the system, NOT
how the changes are made!)
-- cannot measure the enthalpy of a system but the change in enthalpy = ∆H
-- change in enthalpy ∆H = Hfinal – Hinitial = qp
where p reminds you that heat is measured at a constant pressure
-- positive ∆H indicates heat transferred to the system from the surroundings, so
endothermic process.
-- negative ∆H indicates heat transferred from the system to the surroundings, so
exothermic process.
Note: In chemistry, two kinds of work: electrical work and mechanical work. We are
thinking of mechanical work for the most part when we are talking about internal energy
and enthalpy. With gases the work is pressure-volume work.
IV. Enthalpies of Reaction:
Enthalpy of reaction = ∆Hrxn = H(products) – H(reactants)
The thermochemical equation for the combustion of hydrogen is:
For the reaction
2H2(g) + O2(g)  2H2O(g)
∆H = −483.6 kJ
∆H is at the end of the balanced equation is understood that the coefficients in the balanced
equation are the number of moles of reactants and products associated with this enthalpy
change.
Enthalpy of a system = the PE of the system or its heat content.
Guidelines for using Thermochemical equations and Enthalpy diagrams:
1. Enthalpy is an extensive property. ∆H is directly proportional to the amount of
reactant consumed in the process.
2. The enthalpy change for a reaction is equal in magnitude BUT opposite in sign to ∆H
for the reverse reaction.
Example: CH4(g) + 2O2(g)  CO2(g) + 2H2O(l)
∆H = −890 kJ
CO2(g) + 2H2O(l)  CH4(g) + 2O2(g)
∆H = 890 kJ
3. The enthalpy change for a reaction depends on the state of the reactants and products.
Example: 2H2O(l)  2H2O(g)
∆H = +88 kJ
The above reaction with methane would have ∆H = −802 kJ if water was in the
gaseous state instead of the liquid state since gaseous water has a higher heat
content than the liquid state.
Also, generally assume temperature is at 25ºC.
∆Hrxn can be determined experimentally or calculated from a series of enthalpy changes
associated with other reactions.
How much heat is released when 4.50 g of methane gas is burned in a constant-pressure system? Use data
in notes above!
V. Calorimetry
Heat capacity = the amount of heat required to raise its temperature by 1 K (or 1ºC)
-- larger the heat capacity the more heat required to raise a rise in temperature.
Molar heat capacity = the heat capacity of 1 mole of a substance
Specific heat = heat capacity of 1 gram of a substance
q = (specific heat) x (grams of substance) x ∆T
a couple of notes here…
1. specific heat is usually designated as C (yes, same as nutritional calories
or kilocalories) Specific heats are known for many substances and can be
found easily in tables (watch the units though)
2. physics likes to use kilograms instead of grams, so watch units given with
typically the specific heat
3. ∆ T in K = ∆T in ºC
(a) How much heat is needed to warm 250 g of water (about 1 cup) from 22ºC (about
room temperature) to near its boiling point, 98ºC ? The specific heat of water is
4.18 J/g-K
(b) What is the molar heat capacity of water?
IN CONSTANT-PRESSURE CALORIMETRY, we assume that the calorimeter perfectly
prevents the gain or loss of heat from the solution to its surroundings (in general chemistry
labs, the equipment is rarely good enough, BUT we make the assumption anyway!) So,
THE HEAT PRODUCED BY THE REACTION IS ABSORBED BY THE SOLUTION!
For an exothermic reaction, heat lost by the reaction and gained by the solution.
For an endothermic reaction, heat gained by the reaction is lost by the solution.
We can easily measure the change in temperature of the solution. So, qsoln = −qrxn
For dilute aqueous solutions, assume the qsoln = qwater = 4.18 J/g-K
Keep in mind that if ∆T > 0, then the reaction must be exothermic and q rxn < 0
When a student mixes 50 mL of 1.0 M HCl and 50 mL of 1.0 M NaOH in a calorimeter, the temperature of
the resultant solution increases from 21.0ºC to 27.5ºC. Calculate the enthalpy change for the
reaction, assuming that the calorimeter loses only a negligible quantity of heat, that the total
volume of the solution is 100 mL, that its density is 1.0 g/mL, and that its specific heat is
4.18 J/g-K.
Start first by calculating the mass of aqueous solution, then calculate qrxn. Since the
process occurs at constant pressure, ∆H =qp
We must next put the enthalpy change on a molar basis for HCl or NaOH (in this case both
are the same), the product of the solution volume of the HCl x the Molarity of HCl
IN CONSTANT-VOLUME CALORIMETRY, a reaction is studied using a bomb
calorimeter. The bomb, which can withstand high pressures, contains a small cup in which
to place a sample, an inlet for adding oxygen, and electrical contacts to initiate a combustion
reaction. The bomb is then placed within an insulated calorimeter which contains a
measured amount of water. Current is passed through the ignition wires until the sample
ignites. Heat is absorbed by the water within the calorimeter. The difference in the
temperature of the water before and after the combustion is used with the known heat
capacity of the calorimeter to calculate the heat of combustion of the sample. So that
qrxn = −Ccalormimeter x ∆T
(Note to find Ccalormimeter , a sample that releases a known quantity of heat is combusted.
The accepted value for C for that sample in a bomb is divided by the temperature change
found for that sample in our calorimeter to give the Ccalormimeter for our bomb calorimeter.
We essentially perform a calibration for our bomb.)
Methylhydrazine, CH6N2, is commonly used as a liquid rocket fuel. The combustion
of methylhydrazine with oxygen produces N2(g), CO2(g), and H2O(l):
2CH6N2(l) + 5O2(g)  2N2(g) + 2CO2(g) + 6H2O(g)
When 4.00 g of methyhydrazine is combusted in a bomb calorimeter, the temperature
of the calorimeter increases from 25.00ºC to 39.50ºC. In a separate experiment, the
heat capacity of the calorimeter is measured to be 7.794kJ/ºC. What is the heat of
reaction for the combustion of a mole of CH6N2 in this calorimeter?
First calculate qrxn for the 4.00 gram sample
Convert the kJ to the heat of reaction for one mole of CH6N2.
In a bomb calorimeter, the heat transferred corresponds to the change in internal energy ∆E
rather than the change in enthalpy, ∆E, because it is carried out with constant volume
conditions. But because the difference between ∆E and ∆H is very small, less than .1%, we
do not correct the measured heat changes to obtain the ∆H value (ie, assume ∆E = ∆H)
VI. Hess’s Law:
Hess’s Law states that if a reaction is carried out in a series of steps, ∆H for the reaction will
be equal to the sum of the enthalpy changes for the individual steps.
Remember that enthalpy is a state function so the enthalpy change, ∆H of a chemical
process depends only on the amount of matter that undergoes the change and on the nature
of the initial state of the reactants and the final state of the products.
The combustion of methane can be thought of occurring in two steps:
CH4(g) + 2O2(g)  CO2(g) + 2H2O(g)
2H2O(g)  2H2O(l)
CH4(g) + 2O2(g) + 2H2O(g)  CO2(g) + 2H2O(l) + 2H2O(g)
Net equation:
CH4(g) + 2O2(g)  CO2(g) + 2H2O(l)
∆H = −802 kJ
∆H = −88 kJ
∆H = −890 kJ
∆H = −890 kJ
Hess’s Law provides a means of calculating energy changes that are difficult to measure
directly.
The enthalpy of combustion of C to CO2 is −393.5 kJ/mol C, and the enthalpy of combustion of CO to CO2
is −283.0 kJ/mol CO:
(1)
C(s) + O2(g)  CO2(g)
∆H = −393.5 kJ
(2) CO(g) +½O2(g)  CO2(g)
∆H = −283.0 kJ
Using this data, calculate the enthalpy of combustion of C to CO:
(3)
C(s) + ½O2(g) CO(g)
Calculate ∆H for the reaction
2C(s) + H2(g)  C2H2(g)
Given the following reactions and their respective enthalpy changes:
C2H2(g) + 5/2 O2(g)  2CO2(g) + H2O(l)
∆H = −1299.6 kJ
C(s) + O2(g)  CO2(g)
∆H = −393.5 kJ
H2(g) + ½ O2(g)  H2O(l)
∆H = −285.8 kJ
VII. Enthalpies of Formation:
Enthalpy of formation: process used for tabulating thermochemical data is the formation of
a compound from its constituent elements.
--∆Hf the heat of formation ; indicates substance formed from its
elements
--magnitude of enthalpy change depends on temperature, pressure,
and state (gas, liquid, or solid, crystalline form)
Standard state = set of condition at which most enthalpies tabulated
1 atmosphere and the temperature of interest, usually 298 K (25ºC)
∆Hº where the superscript º indicates standard-state conditions
Standard enthalpy of formation ∆Hfº = change in enthalpy for the reaction that forms 1 mole
of the compound from its elements, with all
substance in their standard states.
-- usually at 298 K and standard atmospheric pressure
If an element exists in more than one form under standard conditions, the most stable form
of the element is used for the formation reaction. (more stable = lower energy)
Examples: H2 not H, 02 not O or O3, C(graphite) not C(diamond) ∆Hfº = 0
Stoichiometry of formation reactions always indicates 1 mole of substance being produced.
Standard enthalpy of formation of the most stable form of any element is zero.
We can calculate the standard enthalpy change for any reaction that we know the ∆Hfº
values for all reactants and products.
C3H8(g) + 5O2  3CO2(g) + 4H2O(l)
(1)
C3H8(g)  3C(s) + 4H2(g)
(2) + 3C(s) + 3O2(g)  3CO2(g)
(3) + 4H2(g) + 2O2(g)  4H2O(l)
C3H8(g) + 5O2(g)  3CO2(g) + 4H2O(l)
∆H1 = −∆Hfº[C3H8(g)]
∆H2 = 3∆Hfº[CO2(g)]
∆H3 = 4∆Hfº[H2O(l)]
∆Hrxnº = ∆H1 + ∆H2 + ∆H3
∆Hrxnº = ∆H1 + ∆H2 + ∆H3
= −∆Hfº[C3H8(g)] + 3∆Hfº[CO2(g)] + 4∆Hfº[H2O(l)]
= −(−103.85 kJ) + 3(−393.5 kJ) + 4(−285.8 kJ) = −2220 kJ
Notice that we assume that the stoichiometric coefficients in the balanced equation
represent moles. Because enthalpy is an extensive property, the enthalpy change is
multiplied by the number of moles for that substance. We may have to reverse the
formation reaction as in step (1).
In general, ∆Hrxnº = ∑ n ∆Hfº(products) − ∑ m ∆Hfº(reactants)
Where n and m are the stoichiometric coefficients of the reaction
(a) Calculate the standard enthalpy change for the combustion of 1 mole of benzene, C6H6(l), to CO2(g)
and H2O(l). (b) Compare the quantity of heat produced by combustion of 1.00 g propane to that
produced by 1.00 g benzene.
(a)1. Write a balanced equation for the combustion of benzene.
(a)2. Calculate ∆Hrxnº for the reaction. (See Table 5.3, page 167 or Appendix C, page 1019)
∆Hfº[CO2(g)] = −393.5 kJ
∆Hfº[H2O(l)] = −285.8 kJ
∆Hfº[C6H6(l)] = 49.0 kJ
∆Hfº[O2] = 0 kJ
(b)1. ∆Hrxnº for 1 mole propane = −2220 kJ. So for benzene and propane, use molecular
masses to convert moles to grams to get heat of combustion per gram of each substance.
VIII. Foods and Fuels:
Fuel value: the energy released when 1 gram of a material is combusted
--fuel values are positive numbers
Body energy comes from carbohydrates and fats.
Carbohydrates decomposed in intestines into glucose C6H12O6.
C6H12O6(s) + 6O2(g)  6CO2(g) + 6H2O(l)
ΔHº = −2803 kJ
The average fuel value of carbohydrates is 17 kJ/g (4 kcal/g)
Fats also produce CO2 and H2O in their metabolism.
The reaction of tristearin, C57H110O6, a typical fat, is
2C57H110O6(s) + 163O2(g)  114CO2(g) + 110H2O(l) ΔHº = −75,520 kJ
The average fuel value of fats is 38 kJ/g (9kcal/g)
Excess energy is stored as fats---(1) insoluble in water and (2) produce more energy per
gram than either proteins or carbohydrates.
Fuels: Combustion of fuels produces CO2 and H2O with large negative enthalpies of
formation. The greater the carbon and hydrogen content in a fuel, the higher its fuel value.
Coal is the most abundant fossil fuel—80% in US, about 90% in world
Syngas—use coal to produce a mixture of gaseous hydrocarbons. The coal is typically
pulverized and treated with superheated steam. Many pollutants can be removed.
Nuclear energy—energy released in the splitting or fusion of the nuclei of atoms.
Used to produce about 22% of US electricity.
Nonrenewable sources of energy—coal and nuclear energy
Renewable sources of energy—solar, wind, geothermal, hydroelectric, biomass energy from
crops